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CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity     Quiz 5 - MCQExams.com

The limit of xsin(e1x) as x0
  • is equal to 0
  • is equal to 1
  • is equal to e2
  • does not exist
The function represented by  the following graph is.

572668_db4e79fb2abd478fa7332735dca6ad0f.png
  • Differentiable but not continuous x=1
  • Neither continuous nor Differentiable at x=1
  • Continuous but not Differentiable at x=1
  • Continuous but Differentiable at x=1
If [x] denotes the greatest integer not exceeding x and if the function f defined by f(x)= \begin{cases}\dfrac{a+2\cos\,x}{x^2}&(x < 0) \\ b\,\tan \dfrac{\pi}{[x+4]}&(x \ge 0) \end{cases} is continuous at x=0, then the order pair (a, b) =
  • (-2, 1)
  • (-2, -1)
  • (-1, \sqrt{3})
  • (-2,-\sqrt{3})
\displaystyle \lim_{x\rightarrow 0}\dfrac {1 - \cos x}{x^{2}} is ____
  • 2
  • 3
  • \dfrac {1}{2}
  • \dfrac {1}{3}
If \displaystyle \lim_{x\rightarrow \infty}\dfrac{x^3+1}{x^2+1}-(ax+b)=2, then
  • a=2 and b=-1
  • a = 1 and b = 1
  • a = 1 and b = -1
  • a = 1 and b = -2
\lim _ { x \rightarrow 0 } \dfrac { ( 27 + x ) ^ { 1 / 3 } - 3 } { 9 - ( 27 + x ) ^ { 2 / 3 } }  equals :
  • - 1 / 6
  • 1 / 6
  • 1 / 3
  • - 1 / 3
The value of the constant \alpha and \beta such that \displaystyle \lim_{x\rightarrow \infty}\left(\displaystyle\frac{x^2+1}{x+1}-\alpha x-\beta\right)=0 are respectively.
  • (1, 1)
  • (-1, 1)
  • (1,-1)
  • (0, 1)
What is \displaystyle \lim_{x \rightarrow 0 }  x^2 \sin \left(\frac{1}{x}\right) equal to ? 
  • 0
  • 1
  • 1/2
  • Limit does not exist.
If the function f(x) satisfies \displaystyle \lim_{x\rightarrow 1}\frac{f(x)-2}{x^2-1}=\pi, then \displaystyle \lim_{x\rightarrow 1}f(x)=
  • 2
  • 3
  • 1
  • 0
The limit of \left[\frac{1}{x^2}+\frac{(2013)^x}{e^x-1}-\frac{1}{e^x-1}\right] as x\rightarrow 0
  • Approaches +\, \infty
  • Approaches - \,\infty
  • Is equal to log_e(2013)
  • Does not exist
\displaystyle \lim_{x\rightarrow 0}\frac{log_e(1+x)}{3^x-1}= ____.
  • log_e3
  • 0
  • log_3 e
  • 1
The limit of \displaystyle \sum_{n=1}^{1000}(-1)^nx^n as x\rightarrow \infty
  • does not exist
  • exists and equals to 0
  • exists and approaches +\infty
  • exists and approaches -\infty
Which one of the following statements is correct?
  • \displaystyle \lim_{x \rightarrow 0} (fog) (x) exists.
  • \displaystyle \lim_{x \rightarrow 0} (gof) (x) exists.
  • \displaystyle \lim_{x \rightarrow 0-} (fog) (x) = \displaystyle \lim_{x \rightarrow 0-} (gof) (x)
  • \displaystyle \lim_{x \rightarrow 0+} (fog) (x) =\displaystyle \lim_{x \rightarrow 0-} (gof) (x)
The limit of \left\{\frac{1}{x}\sqrt{1-x}-\sqrt{1+\frac{1}{x^2}}\right\} as x\rightarrow 0
  • Does not exist
  • Is equal to \frac{1}{2}
  • Is equal to 0
  • Is equal to 1
\displaystyle\lim_{x\rightarrow\frac{\pi}{6}}\frac{\sin\left(x-\displaystyle\frac{\pi}{6}\right)}{\sqrt{3-2cos x}} is equal to :
  • 0
  • \displaystyle\frac{1}{(\sqrt{3}-2)}
  • 1
  • \infty
Evaluate: \displaystyle\lim_{x\to 10}\dfrac{x^2-100}{x-10}
  • 10
  • -5
  •  20
  • 5
 \underset { x\rightarrow 0 }{ lim } \cfrac { { \left( 25 \right)  }^{ x }-2\left( 15 \right)^ x+{ 9 }^{ x } }{ cos6x-cos2x }  is equal to :
  • log \left ( \cfrac {5} {3} \right )
  • \cfrac {1} {4} log 15
  • - \cfrac {1} {16} \left( \cfrac {5} {3} \right) ^2
  • log \left ( \cfrac {3} {5} \right )
\displaystyle \lim_{x\rightarrow \pi/4} \dfrac {\tan x - 1}{\cos 2x} is equal to
  • 1
  • 0
  • -2
  • -1
If \underset{x\to 0}{\lim}\dfrac{x^a\sin^b x}{\sin(x^c)}, a, b, c, \in R \sim \{0\} exists and has non-zero value, then 
  • a,b,c are in A.P
  • a,b,c are in G.P
  • a,b,c are in H.P
  • none of these
\displaystyle \lim_{x\rightarrow 3} = \dfrac {\sqrt {x} -\sqrt {3}}{\sqrt {x^{2} - 9}} is equal to
  • 1
  • 3
  • \sqrt {3}
  • -\sqrt {3}
  • 0
The value of \displaystyle \lim _{ x\rightarrow \pi /6 }{ \cfrac { 2\sin ^{ 2 }{ x } +\sin { x } -1 }{ 2\sin ^{ 2 }{ x } -3\sin { x } -1 }  } is
  • 3
  • -3
  • 6
  • 0
If f\left( x \right) =\begin{vmatrix} \sin { x }  & \cos { x }  & \tan { x }  \\ { x }^{ 3 } & { x }^{ 2 } & x \\ 2x & 1 & 1 \end{vmatrix}, then \displaystyle\lim _{ x\rightarrow 0 }{ \dfrac { f\left( x \right)  }{ { x }^{ 2 } }  } is
  • -1
  • 3
  • 1
  • Zero
If \displaystyle \lim _{ n\rightarrow \infty  }{ \cfrac { n.{ 3 }^{ n } }{ n{ \left( x-2 \right)  }^{ n }+n.{ 3 }^{ n+1 }-{ 3 }^{ n } }  } =\cfrac { 1 }{ 3 } , then the range of x is (When n\in N)
  • [2,\ 5)
  • \left( 1,\ 5 \right)
  • \left( -1,\ 5 \right)
  • \left( -\infty ,\ \infty \right)
\lim _{ x\rightarrow 3 }{ \left( { x }^{ 3 }-4 \right) /\left( x+1 \right)  } =
  • (4/23)
  • (2/23)
  • (1/8)
  • (23/4)
\displaystyle\lim_{x\rightarrow \frac{\pi}{2}}(\pi - 2x^{\cos x}) is equal to :
  • \pi+2
  • \pi-2
  • e
  • e^{-1}
Determine the value of k for which the following function is continuous at x=3.
f(x)=\dfrac{x^2-9}{x-3}, x \neq 3

f(x)=k, x=3
  • 2
  • 4
  • 6
  • 8
If f(x) = \begin{vmatrix} \cos x& x & 1\\ 2\sin x & x^{2} & 2x\ \\ \tan x & x & 1\end{vmatrix}, then \displaystyle \lim_{x\rightarrow 0} \dfrac {f'(x)}{x}.
  • Exists and is equal to -2
  • Does not exist
  • Exist and is equal to 0
  • Exists and is equal to 2
Which of the following statements are true for the function f(x) defined for -1\leq x \leq 3 in the figure shown?

880039_ce1751dc863440c2a77ba31deb50b922.png
  • \lim_\limits{x\to-1^+} f(x) = 1
  • \lim_\limits{x\to2} f(x) does not exist
  • \lim_\limits{x\to1^-} f(x) = 2
  • \lim_\limits{x\to 0^+} f(x) =\lim_\limits{x\to 0^-} f(x)
\dfrac{\displaystyle \lim_{h \rightarrow 0}(h+1)^2}{\displaystyle \lim_{h\rightarrow 0}(1+h)^{2/h}} is equal to
  • e^1
  • e^{-2}
  • e^2
  • e^{-1}
\mathop {\lim }\limits_{x \to 0} {{(1 - \cos 2x)(3 + \cos x)} \over {x\tan 4x}} is equal to

  • 2
  • 1/2
  • 4
  • 3
Evaluate the limit:
\displaystyle \lim_{x \to 0} \left( \frac{x- \sin x}{x}\right) \sin \left(\frac{1}{x} \right)
  • 1
  • -1
  • Does not exist
  • 0
\lim _{ x\rightarrow { 0 }^{ + } }{ \left( { \left( x\cos { x }  \right)  }^{ x }+{ \left( \cos { x }  \right)  }^{ \frac { 1 }{ \ln { x }  }  }+{ \left( x\sin { x }  \right)  }^{ x } \right)  }  is equal to
  • 2
  • 2+e
  • 2+\dfrac { 1 }{ e }
  • 3
If the function f(x)=\dfrac{e^{x^{2}}-\cos x}{x^{2}} for x \neq 0 continuous at x=0 then f(0)=
  • \dfrac{1}{2}
  • \dfrac{3}{2}
  • 2
  • \dfrac{1}{3}

The function f : R /{0} \rightarrow R given by f(x) = \dfrac{1}{x} - \dfrac{2}{e^{2x} -1} can be made continuous at x=0 by
defining f(0) as 

  • 0
  • 1
  • 2
  • -1
\lim\limits_{x\to 0}\dfrac{e^{\sin x}-1}{x}=
  • 0
  • 1
  • -1
  • none of these
Let {P_n} = \prod\limits_{k = 2}^n {\left( {1 - {1 \over {{}^{{}^{k + 1}}{C_2}}}} \right)} . If \mathop {\lim }\limits_{x \to \infty } {P_n} can be expressed as lowest rational in the form \dfrac { a }{ b }  , then value of (a+b) is __________.
  • 12
  • 4
  • 8
  • 10
If \mathop {\lim }\limits_{x \to 0} {\left( {\cos x + a\sin bx} \right)^{\frac{1}{x}}} = {e^2} then
the possible values of 'a'\& 'b'are: 
  • a = 1,b = 2
  • a = 2,b = 1
  • a = 3,b = 2/3
  • a = 2/3,b = 3
\mathop {\lim}\limits_{x \to \frac{\pi}{2}} \tan x =
  • \infty
  • - \infty
  • 0
  • does not exist
Suppose the function f(x)-f(2x) has the derivative 5 at x=1 and derivative 7 at x=2. The derivative of the function f(x)-f(4x) at x=1, has the value equal to?
  • 19
  • 9
  • 17
  • 14
\lim _{ x\rightarrow 0 }{ \log _{ \left( \tan ^{ 2 }{ x }  \right)  }{ \left( \tan ^{ 2 }{ 2x }  \right) = }  }
  • 1
  • 2
  • \dfrac {1}{2}
  • Does\ not\ exist
The value of \displaystyle \lim_{x\rightarrow 0}\dfrac {\sqrt {1-\cos 2x}}{x} equals
  • 0
  • 1
  • \sqrt {2}
  • Does\ not\ exist
If l = \lim_{x \rightarrow 0} \dfrac{ x(1+ a\ cos x) - b\ sinx}{x^3} is finite, 
 where l \in R, then 
  • (a)(a,b) = (-1,0)
  • (b) a & b are any real numbers
  • (c)l = 0
  • (d) l = \frac{1}{2}
\underset{h \rightarrow 0}{lim} \dfrac{\sqrt{x + h} -\sqrt{x}}{h} is equal to 
  • \sqrt{x}
  • \dfrac{1}{2 \sqrt{x}}
  • 2 \sqrt{x}
  • \dfrac{1}{\sqrt{x}}
Evaluate:
\lim\limits_{x\to 0}(1+ax)^{\dfrac{1}{x}}
  • e^a
  • e^{\dfrac{1}{a}}
  • 1
  • None of these
The value of \lim\limits_{x\to 0}\dfrac{1-\cos x}{x^2}
  • \dfrac{1}{2}
  • \dfrac{1}{4}
  • 2
  • none of these
\displaystyle\mathop {\lim }\limits_{\ x \to 0} \cos \frac{x}{2}\cos \frac{x}{{{2^2}}}\cos \frac{x}{{{2^3}}}......\cos \frac{x}{{{2^n}}} is equal to 
  • 1
  • -1
  • \dfrac{{\sin x}}{x}
  • \dfrac{x}{{\sin \,x}}
\lim\limits_{x\to 0}\dfrac{1-\cos x }{x^2}=
  • 4
  • 2
  • \dfrac{1}{2}
  • 1
\lim _{ x\rightarrow 1 }{ \dfrac { \sqrt { 1-\cos { 2\left( x-1 \right)  }  }  }{ x-1 }  }
  • Exists and it equals \sqrt {2}
  • Exists and it equals -\sqrt {2}
  • Does not exist because x-1\rightarrow 0
  • Does not exist because left hand limit is not equal to right hand limit
\underset{x \to 0}{\lim}\dfrac{\sin [ \cos x]}{1+[\cos x]} is
  • 1
  • 0
  • does not exist
  • 2
\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{4x}} - 1}}{x}
  • 1
  • 3
  • 4
  • 2
0:0:3


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